My Math Forum  

Go Back   My Math Forum > High School Math Forum > Elementary Math

Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion


Thanks Tree10Thanks
Reply
 
LinkBack Thread Tools Display Modes
August 28th, 2015, 03:53 PM   #11
Senior Member
 
Joined: Jan 2014
From: The backwoods of Northern Ontario

Posts: 391
Thanks: 70

V8 Archie, I don't understand how you arrived at your concluding implication in your first response.
Please spell it out for me.
Timios is offline  
 
August 28th, 2015, 04:00 PM   #12
Senior Member
 
Joined: Jan 2014
From: The backwoods of Northern Ontario

Posts: 391
Thanks: 70

V8 Archie, what you did in your second response is NOT what I did in my original post.

I subtracted a sequence from an identical sequence, namely 1/2 + 1/4 + 1/8 ... from
1/2 + 1/4 + 1/8 ...
Timios is offline  
August 28th, 2015, 04:10 PM   #13
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,664
Thanks: 2644

Math Focus: Mainly analysis and algebra
No, you subtracted the first term of one series from the second term of another, the second from the third, the third from the fourth, etc..

Anyway, the first example I gave did subtract equal terms from each other and also returned the wrong answer.
v8archie is online now  
August 28th, 2015, 04:13 PM   #14
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,664
Thanks: 2644

Math Focus: Mainly analysis and algebra
Quote:
Originally Posted by Timios View Post
V8 Archie, I don't understand how you arrived at your concluding implication in your first response.
Please spell it out for me.
Which part is that?
v8archie is online now  
August 28th, 2015, 04:28 PM   #15
Math Team
 
topsquark's Avatar
 
Joined: May 2013
From: The Astral plane

Posts: 2,205
Thanks: 901

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by Timios View Post
I don't know what you are doing here. But if you do the analagous thing that I did, using your S and 2S, you would calculate 2S-S as follows:

2S= 2 + 4 + 6 + 8 + ...
S= 1 + 2 + 3 + 4 + ...

Subtract both sides and you have:

S = 1 + 2 + 3 + 4 + ...
The point (which I think is pretty well hammered down by now) is that algebraic manipulations of non-converging sums don't behave well. Many of them give different answers if the series are re-ordered. In this case:
S = 1 + 2 + 3 + 4 + ...

2S = 2(1 + 2 + 3 + 4 +...) = 2 + 4 + 6 + 8 +...

2S - S = (2 + 4 + 6 + 8 + ...) - (1 + 2 + 3 + 4 +...)

2S - S = -1 + (2 - 2) - 3 + (4 - 4) - 5 + (6 - 6) - 7 +... = -1 - 3 - 5 - 7 - ...

Or you could do it v8archie's way and get S = 1/2.

Always make sure your sums converge, and even then be wary of changing the order of your series. These can be very tricky to work with.

-Dan
Thanks from greg1313
topsquark is offline  
August 29th, 2015, 12:32 PM   #16
Senior Member
 
Joined: Jan 2014
From: The backwoods of Northern Ontario

Posts: 391
Thanks: 70

Quote:
Originally Posted by Timios View Post
We will prove that:
$\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}... = 1$ (exactly)

Proof:
(1) Let n = $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$

Multiply each side of (1) by 2

(2) 2n = $\displaystyle 1 + \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$

Subtract each side of line (1) from each side of line (2)

(3) n = 1
First I multiplied each side of (1) by 2. Is there any problem with the resulting equation (2)?

Secondly, I subtracted n from 2n to get n as the left side of (3).
I still see no problem.

Finally I subtracted $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ from the $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ on the right side of (2). This seems to be the problem for some of you.

As far as I can see, in all your counter examples with non-converging series, none of you has subtracted an identical series as has been done above. For that reason, I do not see them as true counter examples.
Timios is offline  
August 29th, 2015, 01:46 PM   #17
Math Team
 
topsquark's Avatar
 
Joined: May 2013
From: The Astral plane

Posts: 2,205
Thanks: 901

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by Timios View Post
First I multiplied each side of (1) by 2. Is there any problem with the resulting equation (2)?

Secondly, I subtracted n from 2n to get n as the left side of (3).
I still see no problem.

Finally I subtracted $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ from the $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ on the right side of (2). This seems to be the problem for some of you.

As far as I can see, in all your counter examples with non-converging series, none of you has subtracted an identical series as has been done above. For that reason, I do not see them as true counter examples.
It only works for you because the sum of your series converges. The other examples show that for series that do not converge the algebra does not work. We are not saying you are wrong...we are saying that you have to check for convergence before you do any of this.

-Dan
topsquark is offline  
August 29th, 2015, 03:05 PM   #18
Senior Member
 
mrtwhs's Avatar
 
Joined: Feb 2010

Posts: 706
Thanks: 141

$\displaystyle n = \dfrac{3}{2}+\dfrac{9}{4}+\dfrac{27}{8}+ \ldots$

Quote:
Originally Posted by Timios View Post
First I multiplied each side of (1) by 2/3. Is there any problem with the resulting equation (2)?

$\displaystyle n = \dfrac{3}{2}+\dfrac{9}{4}+\dfrac{27}{8}+ \ldots$

$\displaystyle \dfrac{2n}{3} = 1 + \dfrac{3}{2}+\dfrac{9}{4}+\dfrac{27}{8}+ \ldots$

Secondly, I subtracted $\displaystyle n$ from $\displaystyle \dfrac{2n}{3}$ to get $\displaystyle \dfrac{-n}{3}$ as the left side of (3).
I still see no problem.

Finally I subtracted $\displaystyle \dfrac{3}{2}+\dfrac{9}{4}+\dfrac{27}{8}+ \ldots$ from the $\displaystyle \dfrac{3}{2}+\dfrac{9}{4}+\dfrac{27}{8}+ \ldots$ on the right side of (2). This seems to be the problem for some of you.
Following your steps yields $\displaystyle \dfrac{-n}{3}=1$ or $\displaystyle n=-3$.

Put yourself in for a Fields Medal. You've shown that a bunch of positive numbers adds to $\displaystyle -3$.
Thanks from topsquark and Timios
mrtwhs is offline  
August 29th, 2015, 03:18 PM   #19
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,664
Thanks: 2644

Math Focus: Mainly analysis and algebra
Quote:
Originally Posted by Timios View Post
First I multiplied each side of (1) by 2. Is there any problem with the resulting equation (2)?

Secondly, I subtracted n from 2n to get n as the left side of (3).
I still see no problem.

Finally I subtracted $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ from the $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$ on the right side of (2). This seems to be the problem for some of you.

As far as I can see, in all your counter examples with non-converging series, none of you has subtracted an identical series as has been done above. For that reason, I do not see them as true counter examples.
My first example:
$$\begin{aligned}
&& S &= \phantom{-1+} 1 - 1 + 1-1+\cdots \\
&\text{multiply by $-1$} & -S &= -1+ 1 - 1 + 1-1+\cdots \\
&\text{subtract} & 2S &= \phantom{-} 1
\end{aligned}$$
v8archie is online now  
August 29th, 2015, 04:16 PM   #20
Senior Member
 
Joined: Jan 2014
From: The backwoods of Northern Ontario

Posts: 391
Thanks: 70

Thank you, mrtwhs. I am convinced that you have provided a true counter-example.

Your sarcasm is also cute.

Last edited by Timios; August 29th, 2015 at 04:34 PM.
Timios is offline  
Reply

  My Math Forum > High School Math Forum > Elementary Math

Tags
adds, infinite, proof, series



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Infinite Series Mathforum1000 Algebra 4 July 12th, 2012 04:25 AM
Proof of tricky infinite series jl18 Number Theory 1 June 10th, 2012 10:41 AM
Infinite series : 1/r^r Zeefinity Real Analysis 4 August 28th, 2011 11:18 PM
Infinite series : 1/2^(n^2) Zeefinity Real Analysis 3 April 22nd, 2010 09:58 PM
advanced adds calculating texas hold'em ricbernards Advanced Statistics 0 April 24th, 2007 08:07 AM





Copyright © 2019 My Math Forum. All rights reserved.